elementary-number-theory

[sleek cove]

uh

prime power fact

[light flicker]

I feel like you're overthinking this hard

or maybe I'm missing something

[sleek cove]

um

[light flicker]

|| means divides right

just making sure

[sleek cove]

exactly divides

so p divides, but not p to a higher exponent

[light flicker]

yeah, maybe this spoils the answer for you, but can't you just take m = n = p^k?

OH

Okay that's where the confusion is

it probably helps to think mod p^k here

[gilded tinsel]

you can just construct such an m and n

no need to do anything more compicated

[light flicker]

Yeah

[gilded tinsel]

here ill give u a hint

[winter bear]

yeah construction immediately gives an answer here

[gilded tinsel]

write m=pa and n=pb

then what can we say about a+b

[sleek cove]

uh, equal to (m+n)/p, so is a multiple of p?

[gilded tinsel]

we want p^k || m+n=pa+pb=p(a+b)

yes?

look at hw many power of p can divide a+b

[sleek cove]

only 1

[gilded tinsel]

no we want k powers of p to divide m+n

if we only have 1 power of p dividing a+b we would only have 2 power of p dividing m+n

we want k powers

p divides p so we already have 1 power of p dividing m+n

[sleek cove]

yes

[gilded tinsel]

what about the other k-1 power of p?

what must that divide?

[sleek cove]

a + b

[gilded tinsel]

yup

so p^(k-1) || a+b

whats the simplest number n you can think of such taht p^(k-1)||n

[sleek cove]

p^(k-1)

[gilded tinsel]

yup

so we can set a+b=p^(k-1)

can we find such an a and b?

remember keep it simple

[sleek cove]

a = b = p^(k-1) /2

[gilded tinsel]

you also should remember that we want a and b to be integers

as m and n are integers

[sleek cove]

oh ye

uh a = p^k, b = -p

nvm

[gilded tinsel]

u sure?

add them up?

that doesnt equal p^(k-1)

[sleek cove]

i feel like im dumb rn

yeah was thinking mult

..idk

[gilded tinsel]

ok lets keep it simple

the simplest positive integer i can think of is 1

so lets say i pick a=1

[sleek cove]

b = p^(k-1) -1

[gilded tinsel]

yup

now we check

is it positive?

is it integer?

[sleek cove]

its always a non neg int

[gilded tinsel]

great

so that means m and n are positive integers

[sleek cove]

i feel like this is kinda cheaty

[gilded tinsel]

and because a+b=p^(k-1) we have splved the problem

[sleek cove]

yea

[gilded tinsel]

mah math ppl are lazy

whenever we want to prove existence of something we only need to know one instance of it

there may be other more complex examples for m and n but as long as we know one for certain

we can say it exists

[sleek cove]

thank you for the help, i have my solution, or what i think it is, doing it the way i started out, will send after i make it neat ish

and yeah

thanks for the help

[gilded tinsel]

yw

[torn escarp]

||this is a nice problem for the p-adic ultrametric inequality, |p^k-p| = max(|p^k|, |p|) = |p|. So p divides p^k-p and p only once, and of course p^k-p+p=p^k.||

[sleek cove]

this is what i came up with, would this work?

[torn escarp]

oh sorry I thought you had just finished solving the problem with rockpaperscissors

[sleek cove]

i did, but i had already started on this method

and i have no idea what the p adic thing is yet, so its fine

like i already did this

[torn escarp]

yeah I figured it would be incomprehensible anyways I'm just advertising lol

[sleek cove]

just making it neater, so someone could check this

[torn escarp]

I don't follow your solution, there's too much going on and my attention span is too tiny right now

[sleek cove]

um ok lol

[spare tangle]

Hi, I was wondering how to do the second part of this question

[light flicker]

what did you get for the first part?

[spare tangle]

-27

[light flicker]

uh no

You don't understand what they're asking

[spare tangle]

I mean I did Euclid in reverse

with 73, 27

[light flicker]

oh

wow

never mind it actually is -27

I thought you didn't understand

[spare tangle]

No no it actually is -27

xD

[light flicker]

okay, now how can we use this to solve our question?

Maybe, think about what you would do to solve 27x = 11 in the rational numbers

[spare tangle]

I get that -27 . 27 should be the identity element

but I'm not sure what operation we're using

like, what's our "dot" in this case, is it multiplication?

[light flicker]

Yes

[spare tangle]

it's just multiplication?

[light flicker]

I mean that's what 27^{-1} means

[spare tangle]

so if we multiply by the inverse on both sides

it's just scalar multiplication or is it multiplication mod 73?

[light flicker]

the latter

[quick furnace]

-27 . 27

ok wait

you got me good

anyway

it's multiplication in Z_73

aka

multiplication modulo 73

[spare tangle]

but then (-27 x 27 ) mod 73 should be equal to 27, right?

[light flicker]

Why?

Remember how you found -27

[spare tangle]

because -27 is the inverse of 27 under Z73?

nvm

I got confused there

between inverse and identity x_X

Btw, on the RHS of the eqn 27x = 11 (mod 73)
How would you represent the inverse of 27, do you have to write it as -27 (mod 73) * 11 (mod 73)

[light flicker]

or just -27 * 11 (mod 73)

[spare tangle]

So you don't have to explicitly state that you're referring to -27 in Z*_73?

[light flicker]

It's assumed

I guess you can write $\overline{-27}$ if you want

[sterile pumice]

Zopherus:

[spare tangle]

So then do I use the brackets, dot or star symbol?

[light flicker]

uh what

[spare tangle]

as the binary operator

[light flicker]

whatever

[spare tangle]

Okay, thank you

Sorry I'm such a noob and thanks a lot for your patience

[winter bear]

first: $$P( (a,b)=1) = \lim \sum_{k\leq n} \dfrac{1}{n}\dfrac{\phi(k)}{k}$$
next we know $$\phi= \mu \star \dfrac{1}{n} \implies \phi(n) = \dfrac{1}{n} \sum_{d|n} \dfrac{\mu(d)}{d}$$. so the sum is $$\sum_{k\leq n} \dfrac{1}{n}\sum_{d|n} \dfrac{\mu(d)}{d} = \dfrac{1}{n}\sum_{d\leq n} \dfrac{\mu(d)}{d} \lfloor \dfrac{n}{d}\rfloor$$
this approaches $$\sum_{d\in \bN} \dfrac{\mu(d)}{d^2} = \dfrac{1}{\zeta(2)}$$

[sterile pumice]

JohnDoeSmith:

[winter bear]

@sacred junco

[torn escarp]

this is how you reason this out?

I've never seen it this way interesting

[winter bear]

idk i just came up with it lol

[torn escarp]

the way I saw it is, the probability two numbers have a prime in common is 1/p^2 so 1-1/p^2 is the probability they're relatively prime

product over all primes -> 1/zeta(2)

[dawn warren]

ya

[winter bear]

oh thats nice

[dawn warren]

thats the normal way

never seen john's way tbh

[torn escarp]

hey whatever works is good

[winter bear]

yeah oof thats a much simpler way to do this

[torn escarp]

when I first saw that I started thinking that the zeta function should really be 1/zeta(s)

$$\frac{1}{\zeta(s)} = \sum_{n=1}^\infty \frac{\mu(n)}{n^s} = \frac{1}{\sum_{n=1}^\infty \frac{1}{n^s}}$$

[sterile pumice]

Merosity:

[torn escarp]

basically felt like the harmonic mean sort of way of thinking about the probability was more natural

and maybe other things would be more obvious to take this kind of stance but I kinda forgot about it lol

[winter bear]

interesting

[lavish urchin]

what is the mu function here?

[torn escarp]

mobius function

it's (-1)^k for k the number of unique prime divisors of n, if it has any nonunique divisors it's 0.

not the nicest way to say what it actually is though

[lavish urchin]

that's interesting

[torn escarp]

one way it naturally pops out is from doing inclusion/exclusion on things that you're counting with divisors, the other way is if you try to factor the zeta function

[lavish urchin]

oh I knew that description sounded vaguely familiar https://youtu.be/uvMGZb0Suyc

[torn escarp]

cool I've only really heard the name never knew what the conjecture was

[supple furnace]

I’ll note that the probability argument isn’t fully rigorous, but it’s good for intuition

[winter bear]

problem: find all $pq|2^p+2^q$

sol: We have $2^{p-1} \equiv -1 (q)$ and $2^{q-1} \equiv -1 (p)$. Suppose both are odd primes. then $-1$ is a quad residue mod both $p,q$. this means both are $1$ mod $4$. This means $-1$ is now a $4$th power in both, and hence they are both $1$ mod $8$. this argument can be done over and over to show both primes are $1$ mod all $2^n$, forcing them to be 1, contradicting primality.

Wlog $q=2$, then $2p| 2^2+2^p\to p|6 \to p=2,3$. hence (2,2),(2,3) and (3,2) are the only sols

[sterile pumice]

JohnDoeSmith:

[winter bear]

@supple furnace this works right?

[supple furnace]

Yeah

[winter bear]

phew finally, yeah took me way to long(over an hr i think lol)

[supple furnace]

Nice job, quite similar to my sol

[winter bear]

nice, whats yours

[supple furnace]

First assume $p,q$ are odd. We have $2^{p-1} \equiv -1 (q)$ and $2^{q-1} \equiv -1 (p)$. Then WLOG $v_2(p-1) \geq v_2(q-1)$. Then there exists odd $k$ such that $k(p-1)$ is a multiple of $q-1$. However FLT again implies $1 \equiv 2^{k(p-1)} \equiv -1 (q)$, contradiction
Otherwise WLOG, $q=2$, then $2p| 2^2+2^p\to p|6 \to p=2,3$. hence (2,2),(2,3) and (3,2) are the only sols

[sterile pumice]

tree3:

[winter bear]

oh thats clean

[supple furnace]

Yeah no QR stuff

But your sol is nice too

[winter bear]

nice problem

[supple furnace]

I feel like our sols are very similar tbh

[solemn agate]

can someone help me with this one

these are the theorems

i did this

[winter bear]

you are almost there

also word of advice, dont use so many symbols like forall or and so many times, it serves to confuse both you and the reader.

(symbolic form that is)

[solemn agate]

oh okaym, this teacher wants us to use symbolic form

hmm

i have no idea what to do next

[winter bear]

well you want to show ab|dc right

[solemn agate]

yes

[winter bear]

you have d= something

you want dc

what should you do

[solemn agate]

hmmm

whats the d= something?

is that the d= (a,b) ?

[winter bear]

using bezuouts that is

d=au+bv

[solemn agate]

oh uhh, i dont think he went over that one yet or unless it has a different name for it?

[winter bear]

dude its on your scratchpaper

you literally wrote it down

[solemn agate]

ohh lmao

okay, so then

[solemn agate]

@winter bear

[winter bear]

got it?

[solemn agate]

noo 😦

am i suppose to substitute the values?

[winter bear]

oh, well how far did you go

remember we want dc right

[solemn agate]

i am on the same spot where i posted my solution

[winter bear]

ok

so you have d=something

and you want dc right

what should you do

[solemn agate]

yes

multply by c on both sides?

[winter bear]

mhm

[solemn agate]

so we have dc= cau +cbv

[winter bear]

right

now you want to show ab|dc

do you see how

[solemn agate]

now we would substitiute c= ap and c= bj?

[winter bear]

yeah go ahead

(substitute appropriately btw, remembering we want ab to come out somehow)

[solemn agate]

dc= ( b j u ) a + (a p u ) b

[winter bear]

ja

good job

[solemn agate]

oh okay so then

[winter bear]

well does ab divide this

[solemn agate]

hmm i think so

ah i dont see it

can you explain please D:

[winter bear]

just see what you wrote

can you factor an ab

[solemn agate]

okay, so we go this

dc = ( au+ bv ) c

[winter bear]

dc= ( b j u ) a + (a p u ) b

this thing you had

is the way to go

[solemn agate]

hmm okay

[winter bear]

do you see an ab here

[solemn agate]

yes

hmm so then

[winter bear]

just factor the ab, then ur done

[supple furnace]

Alternatively you can use FTA since it’s there lol

[solemn agate]

so we will have dc = ac

but what does bju and apv represent?

its just arb integer which we can call it k?

[winter bear]

umm ok ill just show you the sol at this point

you had

$d=au+bv \implies dc=acu+bcv = abju+bapv = ab(ju+pv) \ \implies ab|dc$

[sterile pumice]

JohnDoeSmith:

[winter bear]

you see how this works @solemn agate

[solemn agate]

ah yes i see it

damn, i was making it hard LOL

[sacred junco]

Need to find x^2+x+1 = 0 mod 13^3. I'm using Hensel's lemma and I've checked that the needed constraints are satisfied. So we let f(x) = x^2+x+1 and find 0s mod 13. I've found them to be 3 and 9. Evaluating f and f', f(3)=13 and f'(3)=7 Next, I'm lifting up solutions mod 13^3 by setting y =3+13*inverse(7)=3+13*13*2=29. However, it is not the case that f(29)=0 mod 13^3. What am I doing wrong?

[supple furnace]

You have a sign error; it should be 3-13(inverse 7). And this is mod 13^2, not mod 13^3.

[sacred junco]

I should get -26 instead of -29 with the correct sign right?

Is the inverse mod 13^2?

so if I understand your notes correctly, it should be y=3-13*inverse(7)=3-13*145=-1882 but this doesn't work either

inverse(7) was calculated mod 13^2 as you suggested

[supple furnace]

You get -23 mod 13^2

Which works

[sacred junco]

oh

[supple furnace]

Then lift again after that

[sacred junco]

so the lifting happens iteratively?

damn this is cumbersome lol

[supple furnace]

Yes

It is

[sacred junco]

so if you're asked solving a polynomial mod prime^35

you're basically doomed?

do you have to lift 35 times?

[torn escarp]

actually no, because you get a doubling of power when you lift, so you only have to do about log_2(35) iterations of it

I might be lying, I might be thinking of a specific kind of version

on the other hand there's another version of Hensel's lemma for polynomials that says if you can factor it mod p it lifts to a factorization mod p^k so at least now that you've found

$x^2+x+1=(x-3)(x-9) \mod 13$

[sterile pumice]

Merosity:

[torn escarp]

then you know when you lift it you will have 1=-a-b and 1=a*b so, if a is the solution lifted from 3, you just have to compute b = 1-a mod 13^3 to get the other solution, which is much easier

[sacred junco]

that's nice

I'm lost again. What am I doing wrong this time trying to find -6x=1 mod 7^4? Again immediately x=1 is the solution mod 7 so we check f(1)=-6 and f'(1)=-6. Then y=1-(-6)*inverse(-6)=1+6=7 but f(7)=-6*7=42 is not congruent to 1 mod 7^2

[torn escarp]

what are you using for f, f(x)=-6x-1?

$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$

[sterile pumice]

Merosity:

[torn escarp]

this is what you're doing to iterate yeah?

[sacred junco]

yeah so x_n in this case is 1

f(x_n) in this case is -6

inverse(f'(x_n)) in this case is 1

[torn escarp]

no

[sacred junco]

is it inverse mod 7^2?

[torn escarp]

if f(x)=-6x-1 then f(1) is ?

[sacred junco]

why is f(x)=-6x-1?

[torn escarp]

you need to make sure you put it clearly what f is such that f(x)=0

[sacred junco]

f(x) = -6x

[torn escarp]

you've written your f such that f(x) != 0

which is what's getting you the wrong answer, this gets you the roots of f

[sacred junco]

f(x) = 0 for x = 1

[torn escarp]

ok so you're saying f(x)=-6x is 0 when x=1?

[sacred junco]

ok wait

[torn escarp]

f(1)=-6*1 = -6

[sacred junco]

im getting confused

[torn escarp]

this is why it's wrong

[sacred junco]

hensels lemma needs u to have a derivative that is nonzeo

[torn escarp]

you need to write f(x)=-6x-1=0

because Hensel's lemma gets you the roots of f

[sacred junco]

so x=0

[torn escarp]

if you write f(x)=-6x then the root of this is just x=0

[sacred junco]

is what i should have gone for

okay thank you so much

[torn escarp]

there's actually a way to get the answer instantly and cheat

"cheat"

[sacred junco]

how?

[torn escarp]

$$-6x=1$$
$$x=\frac{1}{-6} = \frac{1}{1-7} = 1+7+7^2+7^3+7^4+\cdots$$

[sterile pumice]

Merosity:

[torn escarp]

so $x= 1+7+7^2+7^3 \mod 7^4$

[sterile pumice]

Merosity:

[sacred junco]

beautiful

[torn escarp]

because higher powers of 7 get chopped off, this is valid

but this is thinking in terms of the 7-adic absolute value

[lavish urchin]

this blows my mind

[sacred junco]

i havent learned p adic numbers yet

[torn escarp]

yeah I'm jus tshowing you that so you can check your answer

by Hensel's lemma

[quick furnace]

did i just see someone attempt to use Newton's method to solve an equation on a discrete domain

[sacred junco]

hensels lemma is related to newtons method

i think u can prove the lemma using newtons method in some parts

[torn escarp]

yeah Hensel's lemma is Newton's method for p-adics

[sacred junco]

that is when you pause and realize how beautiful math is lol

[torn escarp]

😛

[sacred junco]

@torn escarp after practicing with the lemma for a while i've noticed that ur lifting picture is partly not desirable

the division doesnt have to be f'(x_n)

it can be f'(x_0) and it will work too

[torn escarp]

yeah you're right

[sacred junco]

if you're solving these by hand then it is better to leave it as f'(x_0) as to not recompute it every lift

it works either way though

[torn escarp]

yeah I should have probably said that, I wasn't really focusing on that when I put it

[sacred junco]

[torn escarp]

also if you wanna know some more tricks, you also don't need f'(x) != 0 mod p

[sacred junco]

i nominate you to be promoted to @ helpers

[torn escarp]

you just need f(x) to be divisible by a larger power of p than f'(x)^2

for that choice of x to be lifted

[sacred junco]

thats a good shortcut too

[torn escarp]

have you learned the p-adic absolute value yet?

|x|=p^{-n} where n is the largest power of p dividing x

it obeys all the regular things absolute value does and a few more stronger conditions, the reason being it's nicer to write the condition I said a second ago as,

$|f(x)|<|f'(x)|^2$

[sterile pumice]

Merosity:

[torn escarp]

that's the condition for x to have a unique lift

[sacred junco]

nope i haven't gotten into p adics yet

is that still #elementary-number-theory ?

i think p adics in their full glory could go into advanced

[torn escarp]

well if you're hensel lifting I feel like you're already talking about p-adics

but maybe the label sounds scary so they don't say it lol

you gotta start somewhere in anything I guess, like when you first teach the definition of limit in calculus or something, it's not like you're immediately doing hardcore differential geometry so I guess it depends on where you decide to draw the line

[sacred junco]

yeah definitely

[torn escarp]

I guess it seems mean to not teach the p-adics exist from my perspective, you're like

forced to dance around "can I divide?" cause working mod p^n is like that

but p-adics basically say, hey, you're free to divide and there's not really any issue, and then you just take the result and can just round it off to get your answer mod p^n

[sacred junco]

i think i'm following and actually i can just read about it myself at this point if it's not something we continue doing in class

[torn escarp]

yeah I have not given the most precise introduction to it by any stretch lol

feel free to ask if you're interested in this stuff, I will explain whatever this is cool to me

[sacred junco]

[torn escarp]

one thing p-adics have that's constantly getting use is the ultrametric inequality

normally you have the triangle inequality, |x+y| <= |x| + |y|

the p-adic absolute value obeys a stronger version of it,

|x+y| <= max(|x|,|y|)

and even further that statement alone is enough to imply if |x| != |y| then |x+y| = max(|x|, |y|)

and this ends up saying stuff going back to |f(x)|<|f'(x)|^2 I said a moment earlier about what solutions can be lifted

anywho I'll stop rambling there lol

[sacred junco]

it's a bit outside of my comfort zone now

[solemn agate]

for number 4 does he want us to prove the backward direction only?

[sacred junco]

i wouldnt say so

i think you are asked to prove both backward and forward and you just happen to have a hint for backward for whatever reason

[solemn agate]

oh okay

can you help me with the forwward direction

[shrewd marsh]

https://gyazo.com/cd503e5920d76485a9759f39ff4e55f4

https://gyazo.com/12d831046fd6a21a57408fbab35faa20

Why exactly is the last inequality in picture 2 nonsense?

Author says that it's impossible, but I'm confused why

Ohhh sorry...

I got it now

Don't need help anymore

integer between two consecutive integers is obviously nonsense

[pine marsh]

okay quick question... is it true that for some ax congruent b (mod m), where a and b can be any number, the solution to x must be equal to or less than m....?

I believe that's the case but I want to double check

(and by any number, I mean integers)

(and m can be any integer too)

(well m has to be positive but you get what I mean)

[light flicker]

no, you can check that 1 * 3 = 1 (mod 2)

[pine marsh]

okay

because I'm having trouble with:
x congruent 31 (mod 41)
x congruent 59 (mod 26)

find x

I set it up so that I got x = 31 + 41y for the first equation

subbing in for x on the second equation (31 + 41y = 59 (mod 26)) and solving for y yields 22

going back to the x equation and solving for x yields 933

but 933 is the solution to only ONE of the equations

what am I doing wrong?

[light flicker]

I'm not sure you solved for y correctly

[pine marsh]

and also, checking for the number of solutions for 41y congruent 18 (mod 26) shows there's only one solution

how do I solve for y then?

[light flicker]

I mean, how did you solve for it, because that solution doesn't work

[pine marsh]

I think I'm solving for y wrong incorrectly as well

I figured the solution to y couldn't be anything greater than 26

so I checked y for all values in (41y-28)/26, where the outcome must be an int

there's another way of solving this I know

but I couldn't figure it out

also I realized my first value of y is wrong

but I fixed it and I still don't have the correct value

nevermind I'm just going to try and get help

but there's very few help sources for elementary number theory at my campus

actually let me rephrase my question

@light flicker if you have x congruent b (mod m), x cannot be greater than m, right?

[light flicker]

Again, you can just do the same thing

3 = 1 (mod 2)

[pine marsh]

so if that method doesn't work, why do some places tell you try to values of X up until m?

because I think I was put off by some of these online tutorials

[light flicker]

Because there's always something less than m that works

in this case 1 = 1 (mod 2)

[pine marsh]

so I went back to another problem, a more simple one, and I'm trying to solve it with the division method

3x = -1 (mod 2)

I really don't understand euclid's algorithm (division algorithm)

like with the example I'm following (17x congruent 3 (mod 29)), they start by setting it as 29 = 1 * 17 + 12

where did the 1 come from? where did the 12 come from?

[light flicker]

You're dividing 29 by 17

how many times can you put 17 into 29? and what's the remainder?

[pine marsh]

you can put it in 1 time with a remainder of 12

okay that explains that

okay I think I get that now; at least the part where they get it all the way down to 5 = 2 x 2 + 1

[prisma flame]

Can someone explain how to use chinese remainder theorem to solve, a^2 congruent 1 (mod 3^4) and a^2 congruent 1 (mod 5^2)?

I’m confused

[light flicker]

What exactly are you confused about?

[prisma flame]

I get that I need to take the roots to get a congruent to +-1 for both mods

And that I need to use Bezouts/Euclidean theorem to find the linear combination: ax + by... and I found (-4)(81)+(13)(25)

But not sure what to do using that information

@light flicker

[prisma flame]

Basically the question is: "Determine all a in {0,1,2,...,2024} with a_bar^2 = 1_bar in Z/2025Z

<@&286206848099549185>

[sacred junco]

a_bar^2?

Is that just

[sterile pumice]

rudy:

[sacred junco]

@prisma flame

[prisma flame]

Sorry, it's supposed to be (abar)^2

[sacred junco]

What is bar lol

[prisma flame]

@sacred junco

[sacred junco]

As in

a x b x a x r ?

a^2br???

oh

You mean

The overline on top of a?

[prisma flame]

Yes

How do I write that?

[sacred junco]

Determine all a in {0, ..., 2024} with $ (\bar a)^2 = 1 \in Z/2025Z$

[prisma flame]

Okay yes! Thanks!

$ a\bar $

[sterile pumice]

Mac:
Compile Error! Click the reaction for details. (You may edit your message)

[prisma flame]

Oh $\bar a$

[sterile pumice]

Mac:

rudy:

[sacred junco]

This, then?

[sturdy dirge]

$A=\left{x\in\mathbb{Z}/2025\mathbb{Z}|x^2=1\right}.$

[prisma flame]

Yup, except 0,..., 2024 is in the set

[prisma flame]

Thank you 🙂

[sturdy dirge]

The easiest to find are 1 and -1 (aka 2024), now you need to check if there are others.

[winter bear]

This will have 4 roots. Thats prolly not best

Take mod 3^4 and mod 5^2

And then combine the sol

[sacred junco]

,w prime factorize 2025

[sterile pumice]

Display results online and refine query

Requested by rudy#0446

[prisma flame]

Yup, I found that 1 is a solution, however I can't find the others. And I found that the linear combination of $gcd(3^4, 5^2) = 1$ is $1 = (-4)(81)+(13)(25)$ By Bezout's and Euclidean Theorem.

[sterile pumice]

Mac:

[prisma flame]

But I'm stuck on how to apply this to the Chinese Remainder Theorem and find the other solutions

[winter bear]

Combine n=1 mod 3^4 and -1 mod 5^2

For example

Do you know h9w to do this?

[sacred junco]

n = 1 mod p^c has 2 solutions right

For all prime p

[winter bear]

Except 2

Yeah

[sacred junco]

And all pos. integer c

So we need

4 solutions total

[winter bear]

Ill let u takeover rusy. Am on phone

Rudy

[sacred junco]

i am too rip

Hm

[winter bear]

Oof

[prisma flame]

No worries, I can wait for you haha

[sacred junco]

Im not actually sure how to solve this entirely, im thinking her

Yes youre just

Going to have to solve

x^2 = 1 mod 3

And then use Hansel’s lemma I think

I don’t think there’s a very fast way to do this

,w Hensel’s lemma

[prisma flame]

I need to solve the system of congruencies

I'm not sure if I'm allowed to use that because we haven't learned it yet

But I guess I'm not proving anything so its fine\

[sacred junco]

Hensel’s lemma just allows you to scale from mod p to mod p^2 and then to mod p^3 and so on

[prisma flame]

Oh okay

[winter bear]

Ill be on laptop soon

[sacred junco]

https://math.stackexchange.com/questions/564683/how-to-use-the-method-of-hensel-lifting-to-solve-x2-x-1-equiv-0-pmod-11

Oh this guy does something similar here

Perhaps John has a more intuitive solve

[winter bear]

ok so first use some elementery facts

like (Z/p^nZ)* is cyclic for prime p right

so x^2=1 has only 2 sols, +-1

take mods 3^4 and 5^2 as suggested

$x= \pm 1 \pmod{3^4}, x= \pm 1 \pmod{5^2}$

[sterile pumice]

JohnDoeSmith:

[prisma flame]

Yup

[winter bear]

each pair will yield a differenc sol after combining

[prisma flame]

I got that

[winter bear]

now ill show u how to combine

[prisma flame]

And then I have to solve the system of congruencies correct?

[winter bear]

yeah

so suppose x=1 mod(3^4) and -1 mod(5^2) for example

then x= 81n+1 for some n, and we have 81n+1=-1 mod(5^2) and u solve for n mod 5^2 here, and then plug it back in

(you know the inverse of 81 in mod 25, as u did the euclidean algo)

[prisma flame]

hold on trying to see for myself as well

Okay so i'm at 81n = -2(mod 25)

Then -2 (mod 25) is the same as 23(mod 25)

But that doesnt make it easier for me so I should change it to a better number

[winter bear]

well whats the inverse of 81 mod 25

(remember your bezouts identity, what number did you have)

[prisma flame]

(-4) and (13)

[winter bear]

mhm, so whats the inverse

if 1= -4(81)+13 (25)

[take mod 25 of this]

[prisma flame]

-4(81)?

Not sure

[winter bear]

ok let me write something general for you

suppose $x,y$ are coprime, then there is some $a,b$ such that $ax+by=1$. taking mod x, $by\equiv 1 \pmod{x}$ and taking mod y $ax \equiv 1 \pmod{y}$

[sterile pumice]

JohnDoeSmith:

[winter bear]

hence b is inverse of y mod x

and a inverse of x mod y

[prisma flame]

Trying to find if it mentions this in my book

[winter bear]

hmm ok, i mean im just showing you a way to find inverse

you can just try guessing to find it for 81=6 mod 25

[prisma flame]

Yea, I just wanted to find it so that I can refer to it later on if I get lost again

[winter bear]

fair nuff

[prisma flame]

Hold on trying to wrap my head around it

Okay so first I would take $(-4)(81) \equiv 1 \pmod{25}$?

[sterile pumice]

Mac:

[winter bear]

mhm

[prisma flame]

And then solve?

Okay

[winter bear]

we would say that

$81^{-1} \equiv -4 \pmod{25}$

[sterile pumice]

JohnDoeSmith:

[winter bear]

given this, what can you say about 81n=-2 mod(25)

[prisma flame]

Wait how did you get -4 (mod 25) ?

[winter bear]

remember we said (-4)81=1 mod (25)

if ab=1 mod n, then a is said to be the inverse of b mod n, and the inverse is denoted by negative power ofc

[prisma flame]

Yea but it seems like we just took the -4 from the one side of the congruency and multiplied with the 1 to get -4

[winter bear]

i mean i just flipped the sides

-4=81^(-1)

if that makes it better

[prisma flame]

So $(-4)(81) \equiv 1 \pmod{25}$ is the same thing as $81^-1 \equiv -4 \pmod{25}$?

[winter bear]

yes

[sterile pumice]

Mac:

[prisma flame]

Ahhh okay

[winter bear]

its how we define inverses

so given that, lets go back

81n=-2 yeah

[prisma flame]

Yup

[winter bear]

(mod 25)

what do you think we should do

to solve for n

[prisma flame]

Do we swap the 81?

Hmm

[winter bear]

well we know the inverse of 81

use that

[prisma flame]

We can make the -2 into -4?

[winter bear]

multiply both sides by the inverse

what happens

[prisma flame]

Ah the 81 disappears

[winter bear]

mhm

so whats the sol for n mod 25

[prisma flame]

I got n = 8 mod 25

[winter bear]

exactly

[prisma flame]

Nice!

[winter bear]

so 8=25k+8 yes

n=25k+8

for some k

and we said x=81n+1

combine these

and you will get your solution mod 2025

try doing the other case to practice (-1 mod 81 and 1 mod 25)

[prisma flame]

Okay so I got 649 mod 2025

Is that correct?

[winter bear]

yeah

[prisma flame]

Nice!

Thanks so much!

[winter bear]

649^2 %2025

,calc 649^2%2025

[sterile pumice]

Result:

1

[prisma flame]

Sorry for making you pull your hair out while explaining to me

[winter bear]

nah its fine

now try finding the 4th root

its p much the same, but will be good practice

[prisma flame]

Okay

For (-1 mod 81 and 1 mod 25) I got -649 (mod 2025)

[winter bear]

yep

[prisma flame]

I see now why this makes sense

[winter bear]

(notice u didnt actually need to do it again, you could have just negated)

i just made you do it for the practice

[prisma flame]

Yes! Makes sense\

Im doing it for n = -1 for both mods

Let's say I take n = 81 k - 1 and I plug it giving 81k - 1 = -1 (mod 25)

Then I get 81 k = 0 mod 25

But that just means k = 0 mod 25

So I get -1 mod 2025

[winter bear]

if both of them are the same mod, then the solution is obvious, you dont need to combine em like this

i.e both 1 means the combined is 1

and both -1 means combined is -1

[prisma flame]

Oh true, I didnt even see that

and I guess I should change the answer from -1 to 2026

[winter bear]

2024

[prisma flame]

Yea 2024

Since we dont want negatives?

[winter bear]

i mean doesnt matter

-1 is often more convinient

so its better left like that

[prisma flame]

Okay

Thanks a lot @winter bear

Are you doing your masters/Phd? @winter bear

[winter bear]

im in hs

[prisma flame]

Okay

Cool

Thanks again

[winter bear]

np

[dreamy cloud]

$\begin{cases} x \equiv 1 (mod 8) \ x \equiv 3 (mod 9) \ x \equiv 9 (mod 12) \ x \equiv 3 (mod 15) \end{cases}$

[sterile pumice]

Sacha:

[dreamy cloud]

How do I change it to make the mods relatively primes, so I can use the Chinese remainder theorem ?

[light flicker]

what

?????????

[sacred junco]

rip yep i

didnt set correct remainder

[light flicker]

none of that was correct lmao

[sacred junco]

yeah lmao i have to

do the arithmetic rip forgot rules

[stuck lintel]

Did something get deleted?

[sacred junco]

Ye i sent a bunch of everything wrong

[stuck lintel]

Anyways @dreamy cloud you can basically just separate it into coprime mods

[sacred junco]

yes but

He has to fix

The remainder

[stuck lintel]

Like x=9 (mod 12) -> x=9=1 (mod 4), x=9=0 (mod 3)

[sacred junco]

Yes, 12-3=9 so x=9

[dreamy cloud]

oh I see

tysm 🥰

[sleek cove]

what does tha bracket notation mean here?

wait is it just the residue class

mod n

[light flicker]

well you're doing intersection so

its the set of all elements equal to a mod n

[sleek cove]

yea

this should be enough, right

[light flicker]

yeah

[sacred junco]

hey guys, does any of you have notes for studying congruences by Niven, Zuckerman?

[sleek cove]

is it true that for all ax = c mod n, n prime, a = n-1, a is its own multiplicative inverse?

for k = n-2

im pretty sure it is, but not sure how to prove it

a(a^-1) - 1 = nk

like getting a^-1 = n-1 =a is easy if can assume k = n-2, but not sure how to rigorously get that

[daring ice]

Find the set of integer solutions to $24x^5-y^2+5=0$

[sterile pumice]

Token:

[daring ice]

I tried looking at $mod 5$, I get $4x^5+4y^2\equiv 0$ mod $5$, implies $4(x+y^2)\equiv 0$ mod $5$

[sterile pumice]

Token:

[daring ice]

So $y\equiv 0,1,4$ and then $x\equiv 0, 4,1$? Is that all solutions?

[sterile pumice]

Token:

[light flicker]

@sleek cove you can check that (n-1)(n-1) = n^2 - 2n + 1 is 1 mod n

@daring ice that logic doesn't work, when you reduce mod 5, you're losing information

Not all solutions of the mod equation will be solutions to your original equation

[daring ice]

ok

[light flicker]

(taking mod is not an injective function)

I mean you can check that x,y = 0 doesn't work for example

[sleek cove]

thanks

[daring ice]

oh right.

But these are all possible solutions mod 5 so does the final number need to be a subset of these solutions?

Or are there solutions I completely missed by taking mod 5?

[light flicker]

Yeah, these are a subset

All the solutions must satisfy these conditions

[daring ice]

My other solutions don't seem to work either though.

x=4, y=1 doesnt work. and y=4, x=1$ doesnt work.

[light flicker]

But some numbers that are 4 or 1 mod 5 might work

That aren't those numbers

[daring ice]

So I should take different modulus to get more accurate solutions?

Like mod 3?

[light flicker]

That's one way

[daring ice]

Or is there a better way to do this?

If I do that and use chinese remainder theorem to find solutions mod 15. Are all of the solutions going to work or do I just get that some of the numbers of that form work?

[light flicker]

It's the same as what I already said

[daring ice]

So doing that won't get me the exact set of solutions

[light flicker]

Nope

[daring ice]

So how do I do that?

[light flicker]

I mean okay, what happens when you reduce down mod 3

You get that y^2 = 2 (mod 3)

So what solutions for y are there

[daring ice]

None

I guess i dont know what to do then is there a correct choice of modulus that makes this work or do I need to do something else?

[light flicker]

I mean, as we said

when you reduce your modulus, your set of solutions is going to be a subset of what you found for the modulus right

[daring ice]

So I cant aolve it with mod since Ill always get a larger set then what I want?

[light flicker]

But what's the set you got when you reduced down mod 3?

[daring ice]

The set of solutions? There's none right because there are no integers which are the square root of 2

[light flicker]

Yes

[daring ice]

So it has no solutions mod 3

So should I pick a modulus which has solutions?

[light flicker]

As you said, your set of solutions is empty for the modulus 3

And as we said, your set of solutions is going to be a subset of what you found for the modulus

[daring ice]

I don't understand.

So I can't find the set of solutions this way.

[light flicker]

Your set of solutions is empty for the modulus 3

[daring ice]

So then there are no solutions?

[light flicker]

your set of solutions overall is a subset of your set of solutions for the modulus 3

Yes

[daring ice]

Are solutions mod n always a subset of solutions for all integers?

Does it matter if n is prime or not?

[light flicker]

Yes

[daring ice]

Wait there are solutions mod $3$ though, because you get $2y^2+2\equiv 0$ so $y=1$ is a solution

[light flicker]

I mean, if you find a solution 24x^5 - y^2 + 5 = 0

then you must have that 24x^5 - y^2 + 5 = 0 (mod n) is true

[daring ice]

oh nvm

[light flicker]

yeah no that's not right

[daring ice]

Why does it have to be true mod n though?

Like why does a solution to the general equation have to give at least as many solutions mod n?

And obviously some modulus will have more then others but why?

Like mod 5 has solutions, but why does it have solutions and not mod 3?

[light flicker]

By what I just said

If you find x,y such that 24x^5 - y^2 + 5 = 0 is true

then it must be true that 24x^5 - y^2 + 5 = 0 (mod n)

[daring ice]

Sure but why does mod 5 have solutions? I must have lost some constraint on the solutions mod 5, but I apparently didn't lose it mod 3?

I don't see if there's like some reason that reducing certain modulus makes more sense then others. I just chose 5 and 3 randomly.

[light flicker]

Not really, its just a fact of the coefficients of the equation which modulus matters and which doesn't

[wet siren]

Hey. Here's my proof for Goldbach Conjecture. (Please don't steal it, I'm posting it here to help you in your number theory journey before my papers get published by Clay)

Let's take some examples for Goldbach :
4 = 2 + 2
6 = 3 + 3
8 = 5 + 3
Computers have shown that we can do the same thing for very huge numbers, which let us state the following :
Let p and q two primes and n an even number bigger than 2.
n = p+q
And because of Fermat's last theorem, we know that raising all the values of this equation to a power bigger than 2 does not have solutions in Z.
Let's consider the function :
f(n) = adds two primes.
This function works for all even numbers. Here is the demonstration :
f(2k) = f(p + q)
And canceling the function in both sides :
2k = p+q
2k > 2 because I personally defined this function with this condition.
Thus, CQFD

That's the short proof though, the other one has more details. But it's the same ideas.
I'm so happy I contribute for math evolution.

[sacred junco]

Thanks mate, I already posted it under my own name on arxiv

[wet siren]

Sorry but it's considered stealing. Nevermind, I've already given this to Annals of Mathematics before you post my proof, so bad for you lmao

[sacred junco]

Lol terry already called me to give me 1 billion dollars laila 😉

[wet siren]

? Well that's the short proof so yeah @sacred junco

@sacred junco Absoluetly not kidding doyou know that he's my cousin?

I can call him lmao

Really jan ?

[sacred junco]

too late the billion is already on my paypal

[winter bear]

jan prove its not a proof

[wet siren]

What the heck it's not possible

You joking lmao

Well yeah

[winter bear]

yes, i dont see whats wrong

[wet siren]

??I defined my function

You give an input of n and the output is an additive expression

[sacred junco]

Looks fine to me as well, thanks a lot mate

[wet siren]

If we give an even number to the function, the output is p+q and p and q are primes

for an input n we receive a+b

Heyy I've been doing calculus for 2 years

Just primes because we've got no precise even number

2k = p + q

You get 13+3

I'm back

What???

[empty nymph]

"i was only pretending to have the brain of a dying whale ok xdd"

[vast dove]

Ive submitted to the annals of mathematics

posts it in math discord
LMAO

[wet siren]

She says to the Goldbach Conjecture killer

Lmao

[light flicker]

So is this serious or

[supple furnace]

Goldbach isn’t even a millennium prize problem

[sacred junco]

Goldbach is a trivial problem that @supple furnace can solve in his sleep

[sweet sentinel]

Solve: 2^x ≡ 100(mod 101), given that 2 is a primitive root modulo 101, and
101 is a prime number

help pls

[stuck lintel]

2^x=-1 (mod 101) so 2^2x = 1 (mod 101). We know that 2^100=1 so x=50 (I think)

[sacred junco]

,w 2^x = 100 mod 101

[sterile pumice]

Display results online and refine query

Requested by rudy#0446

[sacred junco]

@sweet sentinel you sent that yesterday?

and you solved it

wait

fuck me

fuck

whoops

[alpine jasper]

i'm trying to break this into cases
so let first case be a+bw = 0 (1-w)
this implies 1-w divides a+bw
but note a+bw = (a+b) - b(1-w), so i have to show that a+bw divides (a+b)?
which i'm stuck on

[light flicker]

Have you learned anything about norms

[alpine jasper]

yes

hmm

[light flicker]

Use that

[alpine jasper]

alright

thanks

[light flicker]

It works well with division

Also @ number theory is a ping now

Feel free to use it

[alpine jasper]

like, do i consider $\frac{\lambda(a+b\omega)}{\lambda(1-\omega)}$

[sterile pumice]

Publius:

[alpine jasper]

lambda is the norm

[light flicker]

There should be a theorem you learned about how norms work with division

Euclidean domain if you've heard of that term

[alpine jasper]

yes i have

the theorem says something something about remainder smaller than something something

yeah?

i should probably look it up

is there a role for middle school math that i can have

[light flicker]

Yeah, that's important

[alpine jasper]

does $\frac{\lambda(a+b\omega)}{\lambda(1-\omega)}\in\mathbb Z$ implies that $a+b\omega\equiv 0(1-\omega)$?

[sterile pumice]

Publius:

[alpine jasper]

[light flicker]

No

You should go read the euclidean domain statement

[alpine jasper]

this is what i should be looking at right

i think i'm retarded

[light flicker]

Ah actually your earlier statement was correct

But yeah that's what you should be looking at

[alpine jasper]

Okay. Thank you, that's a lotta new info, I'll try to work from here

[alpine jasper]

okay i'm still stuck:(

i let a+bw = c(1-w) + d
i want to show that d = 0
according to the defn of eculidean norm, i have
N(d) < N(1-w)
N(d) < 3
this shows that d can be -1-w, \pm 1, \pm w, 1+w, and 0

[light flicker]

It's strictly less than

[alpine jasper]

the norm i'm using is N(a+bw) = a^2 - ab + b^2

umm

did WA lied to me

[light flicker]

Oh hm

Yeah I'm not sure this question is right

[alpine jasper]

maybe i can prove this?

idk

[light flicker]

That's true and just follows from euclidean division

[alpine jasper]

oh

hmm

this is what one of my classmate wrote:

Consider a+bw = - b(1-w) + (a+b), note that (a+b) in Z (halfway home), then combine with the hint for problem (25)

maybe that'll help..?

[light flicker]

oof im trolling sorry

A lot of the elements you wrote down are equivalent mod 1 - w

[alpine jasper]

wtf

[light flicker]

i.e 1 = w (mod 1 - w)

[alpine jasper]

i concur

but i don't follow where you're going with this

the problem wants me to show that a+bw = \pm 1, 0 (mod 1-w)

[light flicker]

so now you know that w = 1 (mod 1 - w)

[alpine jasper]

oh fuck.

right

hmm

i still don't know how this will help but gimme a few moments ig

yeah i don't get it, can you explain more pls

am i missing some basic congruent rules

[winter bear]

whats 2 mod (1-w)

[light flicker]

I mean 1 + w = -1 (mod 1 - w)

[winter bear]

or erm sorry 3 mod 10w

1-w

[light flicker]

yeah, its important that 3 is 0 mod 1 - w

[alpine jasper]

jesus

i still don't see anything

i can confirm that 3 = 0 (1-w)

[winter bear]

if w=1 mod(1-w), then what is a+bw mod (1-w)

[alpine jasper]

a+b mod (1-w)

[winter bear]

mhm, and what values can a+b be, considering 3=0 mod w

1-w

[alpine jasper]

let me convince myself why a+b first

... 1 sec

[winter bear]

k

[alpine jasper]

alright

i deduced that i'm fucking shit at congruences

[winter bear]

it takes practice

[alpine jasper]

if w=1 mod (1-w), then
a + bw = a + b(1) mod (1-w)

this is legal right

[winter bear]

yeah

[alpine jasper]

alright

mhm, and what values can a+b be, considering 3=0 mod w
couldn't a+b be anything..?

am i suppose to reduce it mod 3

[winter bear]

yeah. a+b could be anything, but we know that 3=0 mod w, so mod w what could a+b be

1-w

not w*

[alpine jasper]

a+b divide by 1-w gives remainder a+b mod 3

yeah?

[winter bear]

yep

[alpine jasper]

okay

[winter bear]

theres a generalization of this when lambda(z)=p or p^2 in Z[w] btw

[alpine jasper]

$a+b\omega \equiv (a+b \mod(3)) \mod(1-\omega)$

[sterile pumice]

Publius:

[alpine jasper]

does this make any sense

[winter bear]

yeah, notation is a bit ugly though

[alpine jasper]

i don't want any generalization rn

[winter bear]

$a+b\equiv c \pmod{3} \implies a+b\omega\equiv c \pmod{1-\omega}$

[sterile pumice]

JohnDoeSmith:

[winter bear]

fair enough

but good job, you did it

[alpine jasper]

um what

oh

i see

burh

thanks

[winter bear]

np, and thank zoph he did most of the work

[alpine jasper]

yeah he's a big chad

[alpine jasper]

i cant see how to arrive at the hint

[alpine jasper]

<@&681259820611141654>

[supple furnace]

Using (1-w)^2=-3w, (1+a(1-w))^3=1+3a(1-w)+3a^2(-3w)+a^3(-3w)(1-w)=1+3(1-w)(a-a^3w) (equivalent to the hint’s condition). It’s enough to show that a^3w-a is divisible by 1-w, but a^3w-a=a^3-a=0 mod 1-w since Z[w]/(1-w) is a three element finite field.

[alpine jasper]

They were right. This tree3 guy is good.

Tysm

[forest mica]

Can anybody help me?

[light flicker]

Use #discrete-math

[forest mica]

Ok

[alpine jasper]

i'm not sure how to continue this, am i even on the right track?

R is the gaussian integers

[stuck lintel]

@forest mica m, n, k, i are integers?

[light flicker]

He's in #discrete-math

@alpine jasper Use the definition of prime

[alpine jasper]

from p^2 = a^2 + b^2?

the defn of prime that i'm given is that p is prime if p | ab implies p | a or p | b

[light flicker]

Using the fact that primeness is equal to irreducible here might help

[alpine jasper]

isn't that what i did when i assume either x or y is a unit

[light flicker]

Oh, I was misled by the contrapositive, that isn't really contrapositive

Oh, okay, I should've read all of what you had

[alpine jasper]

well it will be once i show p is not congruent to 3 mod 4?

p implies q
i'm trying to prove
not q implies not p

[light flicker]

If p is not a prime, then p = xy doesn't imply that x or y is a unit?

p being prime implies that

[alpine jasper]

what the fuck

i'm retarded

[light flicker]

(or well, irreducible, but they're equivalent in this case)

[alpine jasper]

well there goes my entire proof

[light flicker]

Uh, the same proof just works out if you forget the contrapositive part

just write p = xy, then write N(p) = N(x)N(y)

And you're looking to show that either x or y is a unit

N(p) is p^2

[alpine jasper]

i write p = xy where x and y are both not units, so i'm assuming p not prime

let x=a+bi and y=c+di

then apply norm

p^2 = (a^2 + b^2)(c^2 + d^2)
where (a^2 + b^2)!=1
and (c^2 + d^2)!=1

so p = a^2 + b^2

but this is then painful from what i read online

so i kinda want to avoid that

actually i don't know what i'm doing

let me read what you wrote up there

p^2 = (a^2 + b^2)(c^2 + d^2)
so WLOG let (a^2 + b^2) | p^2
so (a^2 + b^2) | p
...?

[light flicker]

No it's not painful

All you have to do is use the fact that squares are always 0 or 1 mod 4

[alpine jasper]

hmm

[light flicker]

uh, that's not how numbers work

[alpine jasper]

jesus

okay

[light flicker]

p^2 divides p^2 but p^2 doesn't divide p

for example

[alpine jasper]

i see

umm

[light flicker]

just write p = a^2 + b^2

and reread your problem

[alpine jasper]

alright

[light flicker]

There are very important facts

that you haven't used at all yet

[alpine jasper]

i'm being dumb and having trouble with logic again

i want to prove P implies Q here

so i want to show that [not Q and P] is false

[light flicker]

There's really no reason to think about contradiction here, you can do this directly

"just write p = xy, then write N(p) = N(x)N(y)
And you're looking to show that either x or y is a unit"

If we show that x or y is a unit, this shows that p is irreducible (and thus prime)

[alpine jasper]

just write p = xy

is this p from Z or from R..?

[light flicker]

R

[alpine jasper]

i'm guessing its Z

oh

ok

[light flicker]

Because we're trying to show that p is prime in R

Wait

p is from Z sorry

x,y are from R

(p is technically also from R)

[alpine jasper]

oh this fucks my mind even more

i'm still failing to see the logic here, can you just lay it out flat for me please

i want to prove p implies q, so what are we doing exactly here?

[light flicker]

We're trying to show that is p is prime in R

Since prime is the same as irreducible, we try to show that p is irreducible in R

The definition of irreducibility says that if p = xy, with x,y in R, then either x or y is a unit

So we write p = xy, and try to show that either x or y is a unit

To show that p is irreducible, and thus prime

[alpine jasper]

i see

alright

okay